Question

In a Mendelian dihybrid cross the number of phenotypic recombinants is 36 in F2 gene- ration. The number of organisms with the genotype that itself accounts for 1/4th of F, generation is 1) 48 2) 24 16 2 3) 18 4) 12​

Answer 1

Answer:48

Explanation:

Because genotype will be 1/4

Answer 2

Answer:

The correct option among the four here is option 2) 24 organisms with the genotype that itself is accounting for the 1/4th of F is formed.

Explanation:

As we already know, in the case of the Dihybrid cross of Mendel, the phenotypic ratio that we usually obtain in the F2 generation is 9:3:3:1. Where the recombinants are the 3:3, that is total of 6 out of 16 offsprings are recombinants and the rest are the parental generation.

There fore taking the total number of offspring to be x, we have

$\frac{6}{16} x=36$

∴$x = 96$

The number of the genotypes that itself is accounting about

1/4th of F is

$\frac{1}{4} x=\frac{1}{4} of 96$ = 24

∴ The correct option showing the same result as we have got is option

2) 24